| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2012 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Perpendicular bisector of segment |
| Difficulty | Standard +0.3 This is a multi-part coordinate geometry question requiring standard techniques: finding midpoint, gradient of perpendicular line, and properties of squares. While it has multiple steps (7 marks total), each individual step uses routine AS-level methods with no novel insight required. Part (iii) requires recognizing square properties but follows logically from earlier parts. Slightly above average due to length and the square construction, but well within typical P1 scope. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships1.10e Position vectors: and displacement1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(M = (1, 4)\), gradient \(= \frac{1}{2}\) soi | B1B1 | |
| grad of \(MB = -2\) soi | M1 | Use of \(m_1m_2 = -1\) |
| Equation \(MB: y - 4 = -2(x-1)\) | A1 [follow through] | Or \(y = -2x + 6\); ft on *their* \(\frac{1}{2}\) or \(M\) |
| When \(y = 0\), \(x = 3\) so \(B = (3, 0)\) | A1 [follow through] [5] | ft result of putting \(y = 0\) into *their* eqn |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| grad of \(AB = -\frac{2}{6}\); grad of \(BC = \frac{6}{2}\) oe | M1 [follow through] | At least one correct [follow through] |
| \(m_1m_2 = -1 (\Rightarrow AB \perp AC)\) | A1 [2] | AG. Allow omitted conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(D = (-1, 8)\) | B1 | |
| \(AD = \sqrt{40}\) or \(6.32\) | B1 [2] |
## Question 9:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $M = (1, 4)$, gradient $= \frac{1}{2}$ soi | B1B1 | |
| grad of $MB = -2$ soi | M1 | Use of $m_1m_2 = -1$ |
| Equation $MB: y - 4 = -2(x-1)$ | A1 [follow through] | Or $y = -2x + 6$; ft on *their* $\frac{1}{2}$ or $M$ |
| When $y = 0$, $x = 3$ so $B = (3, 0)$ | A1 [follow through] [5] | ft result of putting $y = 0$ into *their* eqn |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| grad of $AB = -\frac{2}{6}$; grad of $BC = \frac{6}{2}$ oe | M1 [follow through] | At least one correct [follow through] |
| $m_1m_2 = -1 (\Rightarrow AB \perp AC)$ | A1 [2] | AG. Allow omitted conclusion |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $D = (-1, 8)$ | B1 | |
| $AD = \sqrt{40}$ or $6.32$ | B1 [2] | |
---9 The coordinates of $A$ are $( - 3,2 )$ and the coordinates of $C$ are (5,6). The mid-point of $A C$ is $M$ and the perpendicular bisector of $A C$ cuts the $x$-axis at $B$.\\
(i) Find the equation of $M B$ and the coordinates of $B$.\\
(ii) Show that $A B$ is perpendicular to $B C$.\\
(iii) Given that $A B C D$ is a square, find the coordinates of $D$ and the length of $A D$.
\hfill \mbox{\textit{CAIE P1 2012 Q9 [9]}}